4 ms·
I found this description disjointed and difficult to read.
by theVirginian 11y ago
I found this description disjointed and difficult to read.
- mpu 11y agoI'm not a native speaker, I'm sorry you had trouble going through it. I tried to keep it short but maybe did not articulate the first paragraph and the end of the second ideally. Hopefully, you understood the core idea.
- mkehrt 11y agoI thought it was clear and concise, but I am familiar with both concepts.
- triplesec 11y agominor edit for you since you're not English: it's Russell's Paradox. It's amazing how much going along with accepted spelling conventions helps people like a piece, especially a short one. Good luck!
- theVirginian 11y agoSorry to be negative, I did get through it but the end of the second was where I had to stop for a second.
- banachtarski 11y agoYou were potentially not in the intended audience.
- tel 11y agoIt might be valuable to expand the definitions of things a bit. First, we talk about characteristic functions. As noted, they are merely functions which return either 1 or 0, True or False. They represent sets by examining their argument and returning True if and only if the set they represent contains that thing. So, if x is a characteristic function and (x x) is True, then x "contains" itself. Then N is just negation, so if N (x x) is True then x does not "contain" itself. That much is simple. Now we form a special new characteristic function, R, defined as R x = N (x x) When is (R x) True? Exactly when x is "contained" in R. When does that happen? When N (x x), e.g., when x does not "contain" itself. So this is a well-defined characteristic function and therefore represents what we might think is a well-defined set. But what happens when we check (R R), i.e. to see if R contains itself? Well, R, by definition, contains all things which are not members of themselves. If R is in R then (R R) is True, but then N (R R) is False and (R R) = False. Contradiction! We can inline R into (R R) in order to take a look at the functional form. (\x -> N (x x)) (\x -> N (x x)) and the author notes that this is the same form as the Y-combinator with its first argument specialized to N. Y f = (\x -> f (x x)) (\x -> f (x x)) Y N = R R And there we go!
- srtjstjsj 11y agoHow is this example special, compared to the general notion: F = (\x -> Z (x x)) Y f = (\x -> f (x x)) (\x -> f (x x)) Y Z = F F so, any set F that is defined in terms of how sets (functions) relate to themselves, fits into the Y combinator. Which is expected, since Y is the essence of recursion.
- tel 11y agoI don't know that there's anything special about Russell's paradox in particular, to be honest. It has obvious historical importance, obviously, but I doubt that's what you're looking for. There's something a bit remarkable about how trivial it is, I suppose. Booleans are very simple types, if you'll give me those, and then Not is the only involution!
- jameshart 11y agoSo to translate that back into natural language, R R asks 'does the set of sets which don't contain themselves contain itself?', and the answer, 'Y N' reads, 'why not'?
- theVirginian 11y agoAhh.. thank you. We touched on this in philosophy courses but my specific courses did not cover it in depth. Thank you for explaining the notation for me.